JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(\omega=z \bar{z}+k_1 z+k_2 i z+\lambda(1+i), k_1, k_2 \in R\). Let \(\operatorname{Re}(\omega)=0\) be the circle \(C\) of radius 1 in the first quadrant touching the line \(y=1\) and the \(y\)-axis. If the curve \(\operatorname{Im}(\omega)=0\) intersects \(C\) at \(A\) and \(B\), then \(30(A B)^2\) is equal to \(.......\).
- A \(23\)
- B \(22\)
- C \(24\)
- D \(21\)
Answer & Solution
Correct Answer
(C) \(24\)
Step-by-step Solution
Detailed explanation
\(\omega=z \bar{z}+k_1 z+k_2 i z+\lambda(1+i)\) \(\operatorname{Re}(w)=x^2+y^2+k_1 x-k_2 y+\lambda=0\) \(\text { Centre } \equiv\left(\frac{-k_1}{2}, \frac{k_2}{2}\right) \equiv(1,2)\) \(\Rightarrow k_1=-2, k_2=4\) \(\text { radius }=1 \Rightarrow \lambda=4\)…
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